Hydraulic losses in pipes

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1 Hydraulic losses in pipes Henryk Kudela Contents 1 Viscous flows in pipes Moody Chart Types of Fluid Flow Problems Minor losses Viscous flows in pipes Our intension here is generalized the one-dimensional Bernoulli equation for viscous flow. When the viscosity of the fluid is taken into account total energy head H = v2 + p ρg + z is no longer constant along the pipe. In direction of flow, due to friction cause by viscosity of the fluid we have v2 1 + p 1 ρg + z 1 > v2 2 + p 2 ρg + z 2. So to restore the equality we must add some scalar quantity to the right side of this inequality 1 + p 1 ρg + z 1 = v2 2 + p 2 ρg + z 2 + h ls (1) This scalar quantity ls is called as hydraulic loss. The hydraulic loss between two different cross section along the pipe is equal to the difference of total energy for this cross section: h ls = H 1 H 2 (2) We must remember that always H 1 > H 2. In horizontal pipe when z 1 = z 2 and diameter of pipe is constant v 1 = hydraulic loss is equal to the head of pressure drop or head loss h L = p 1 p 2 ρg (3) Head loss is express by Darcy -Weisbach equation: h L = f L D (4) 1

2 Figure 1: Pipe friction loss. For horizontal pipe, with constant diameter this loss may be measured by height of the pressure drop: p ρg = h We must remember that equation (4) is valid only for horizontal pipes. In general, with v 1 = but z 1 z 2, the head loss is given p 1 p 2 ρg = (z 2 z 1 )+ f L D (5) Part of the pressure change is due to elevation change and part is due to head loss associated with frictional effects, which are given in terms of the friction factor f that depends on Reynolds number and relative roughness f = ϕ(re, ε/d). It is not easy to determine the functional dependence of the friction factor on the Reynolds number and relative roughness (ε/d). Much of this information is a result of experiments conducted by J. Nikuradse in 1933 and amplified by many others since then. One difficulty lies in the determination of the roughness of the pipe. Nikuradse used artificially roughened pipes produced by gluing sand grains of known size onto pipe walls to produce pipes with sandpaper-type surfaces. In commercially available pipes the roughness is not as uniform and well defined as in the artificially roughened pipes used by Nikuradse. However, it is possible to obtain a measure of the effective relative roughness of typical pipes and thus to obtain the friction factor. Figure (3)) shows the functional dependence of f on Re and and is called the Moody chart in honor of L. F. Moody, who, along with C. F. Colebrook, correlated the original data of Nikuradse in terms of the relative roughness of commercially available pipe materials. 1.1 Moody Chart The following characteristics are observed from the data of (3). For laminar flow, Re < 2300, f = 64/Re, which is independent of relative roughness. For very large Reynolds numbers, f = ϕ(ε/d) which is independent of the Reynolds number. For such flows, commonly termed completely turbulent flow, along the wall pipe, exists the laminar sublayer so thin that the surface roughness completely dominates the character of the flow near the wall. The gap in the figure for which no values of f are given, 2100 < Re < 4000, is a result of the fact that the flow in this transition range 2

3 Figure 2: Flow near rough and smooth walls may be laminar or turbulent (or an unsteady mix of both) depending on the specific circumstances involved. Note that even for smooth pipes the friction factor is not zero. That is, there is a head loss in any pipe, no matter how smooth the surface is made. This is a result of the no-slip boundary condition that requires any fluid to stick to any solid surface it flows over. There is always some microscopic surface roughness that produces the no-slip behavior (and thus f 0) on the molecular level, even when the roughness is considerably less than the viscous sublayer thickness. Such pipes are called hydraulically smooth. Various investigators have attempted to obtain an analytical expression for f = ϕ(re,ε/d). Note that the Moody chart covers an extremely wide range in flow parameters. The non-laminar region covers more than four orders of magnitude in Reynolds number from Re = to Re = Obviously, for a given pipe and fluid, typical values of the average velocity do not cover this range. However, because of the large variety in pipes D, fluids (ρ, and µ and and velocities (v), such a wide range in Re is needed to accommodate nearly all applications of pipe flow. Colebrook combined all data for transition and turbulent flow in smooth as well as rough pipes into the following relation known as Colebrook equation 1 f = 2.0 log ( ε/d Re f The Colebrook equation is implicit in f, and determination of friction factor requires tedious iteration. An approximate explicit relation for f is given by S.E. Haaland in 1983 as [ ( ) ] ε/d 1.11 = 1.8 log f Re + (7) 3.7 ) (6) 3

4 Figure 3: Friction factor as a function of Reynolds number and relative roughness for round pipes -the Moody chart For hydraulically smooth pipe the friction factor is approximated by Blasius (1911) formula f = (100 Re) 1/4 (8) The next formula proposed by Aldsul(1952) gained some popularity in the engineering application due to its simplicity: f = 0,11( ε D + 68 Re )1/4 (9) It is clear that in order to use the Moody diagram we must be able to obtain values of surface roughness. These have been measured and tabulated (and sometimes plotted) for an extensive range of materials used in piping systems. Table 1 provides some representative values. Table 1. Surface roughness values for various engineering materials PIPING MATERIAL ROUGHNESS ε mm Cast iron 0.26 Commercial steel and wrought iron Concrete Drawn tubing Galvanized iron 0.15 Plastic,(and glass) 0.0 (smooth) Riveted steel

5 We must remember that the values in table typically used, are not actual measured ones, but are instead the result of data correlations constructed over a range of measurements. They are sometimes referred to as equivalent roughnesses; it is useful to consider them as simply representative values. 1.2 Types of Fluid Flow Problems In the design and analysis of piping systems that involve the use of the Moody chart, we usually encounter three types of problems: 1. Determining the pressure drop when the the pipe length and diameter are given for a specified flow rate ( or velocity) 2. Determining the flow rate when the the pipe length and diameter are given for a specified pressure drop 3. Determining the pipe diameter when the pipe length and flow rate are given for a specified pressure drop. Example 1. Oil, with ρ = 900kg/m3 and kinematic coefficient of viscosity ν = 0, 00001m2/s, flows at q v = 0,2m3/s through 500 m of 200-mm diameter cast-iron pipe. Determine (a) the head loss and (b) the pressure drop if the pipe slopes down at 10 in the flow direction. Solution. First we compute the Reynolds number Re = V d/ν = 4 q v /(πdν) = 4 0,2/(3,14 0,2 0,00001) = Velocity is equal to v = q v /( πd2 4 ) = 6.4 m/s. Absolute roughness for iron-cast pipe is ε = 0.26 mm. So relative roughness is ε = ε d = 0,0013. Now we are able to calculate using the formula(10) or Moody chart, friction factor f = 0,0225. Then the head loss is h f = f L d For inclined pipe the head loss is So pressure drop is = 0, , ,81 = 117 m h f = p ρg + z 1 z 2 = p ρg + Lsin10o. p = ρg(h f 500 sin 10 o ) = 900 9,81 (117 87) = Example 2. Oil, with ρ = 950 kg/m3 and = 2E 5 m2/s, flows through a 30 cm-diameter pipe 100 m long with a head loss of 8 m. The roughness ratio is ε/d = Find the average velocity and flow rate. Iterative Solution. To start we need to guess f. A good first guess is the fully rough value (wholly turbulent) for ε/d = from Moody chart. It is f Now from Darcy-Weisbach formula (4) we have h f = f L D f =

6 f v = 0.471/0.014 = 5.8 m/s Re = vd/ν f new (87000) = v = 0.471/ = 4.91 m/s Re = vd/ν f new (73700) = v = 0.471/ = 4.84 m/s Re = vd/ν f(72600) = , so we can accept v = 4,84 m/s, q v = v( πd2 4 ) = 0,342 m3 /s. 1.3 Minor losses For any pipe system, in addition to the Moody-type friction loss computed for the length of pipe. Most pipe systems consist of considerably more than straight pipes. These additional components add to the overall head loss of the system. Such losses are generally termed minor losses, with the apparent implication being that the majority of the system loss is associated with the friction in the straight portions of the pipes, the major losses or local losses. In many cases this is true. In other cases the minor losses are greater than the major losses. The minor losses may raised by 1. Pipe entrance or exit 2. Sudden expansion or contraction 3. Bends, elbows, tees, and other fittings 4. Valves,open or partially closed 5. Gradual expansions or contractions The major losses may not be so minor; e.g., a partially closed valve can cause a greater pressure drop than a long pipe. The losses are commonly measured experimentally. The data, especially for valves, are somewhat dependent upon the particular manufacturers design. The most common method used to determine these head losses or pressure drops h L = p ρg = K L (10) where K L means (local) loss coefficient. Although K L is dimensionless, it is not correlated in the literature with the Reynolds number and roughness ratio but rather simply with the raw size of the pipe. Almost all data are reported for turbulent-flow conditions. Example 3. Find (see figure 5.) 1. the discharge through the pipeline as in figure for H=10 m 2. determine the head loss H 1 for q v = 60 l/s. D = 0.15m, ε D = 0,0017, ν = 1, Solution. The energy equation applied between points 1 and 2 including all the losses, can be written as H 1 = v2 + f L D +(K 1 + K 2 + K 3 + K 4 ) v2 6

7 Figure 4: where K 1 = 0,5 is entrance loss coefficient,k 2,K 3 = 0,9 as a standard elbow, and K 4 = 10 for globe valve fully open. Putting the values tot the above formula one obtain H 1 = v2 (13,3+680 f) To start we need to guess f. A good first guess is the fully rough value (wholly turbulent) for ε/d = 0, 0017from Moody chart. It is f 0, 022. Now from Darcy-Weisbach formula (4) we have 10 = v2 (13, ,022) = 2.63, Re = f 0,023 v = 2,6 m/s Re = vd/ν f new (380000) = 0,023 v = 2,6 m/s Re = vd/ν

8 Figure 5: Pipeline with minor losses We accepted v = 2,60 m/s, q v = v( πd2 4 ) = 45, m 3 /s. For the second part, with q v known, the solution is straightforward: = q v A = 0.06 π = 3.40 m/s, Re = , f = and H 1 = 3,42 (13, ,023) = 17,06 m 2 9,81 Example 4. Flow between two reservoirs. Water at 10 o C flow from a large reservoir to a small one through a 5-cm-diameter cast iron piping system shown in figure (6). Determine the elevation z 1 for the flow rate of 6l/s. Figure 6: (Answer. 27,9 m) 8

9 During this course I will be used the following books: References [1] F. M. White, Fluid Mechanics, McGraw-Hill. [2] B. R. Munson, D.F Young and T. H. Okiisshi, Fundamentals of Fluid Mechanics, John Wiley and Sons, Inc.. [3] Y. Nakayama and R.F. Boucher, 1999.Intoduction to Fluid Mechanics, Butterworth Heinemann. [4] Y.A. Cengel and J. M. Cimbala, Fluid Mechanics, McGraw Hill. [5] J.M. McDonough, Lectures in Elementary Fluid Dynamics: Physics, Mathematics and Applications, University of Kentucky, Lexington. 9

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